# What is cos(arccos (3/5) - arcsin (4/5))?

Jul 14, 2018

$1.$

#### Explanation:

Let, $\arcsin \left(\frac{4}{5}\right) = \alpha$.

$\therefore \sin \alpha = \frac{4}{5} , w h e r e , \alpha \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] = {Q}_{1} \cup {Q}_{4}$.

But, since $\sin \alpha = \frac{4}{5} > 0 , \alpha \notin {Q}_{4} \therefore \alpha \in {Q}_{1}$.

Now, ${\cos}^{2} \alpha = 1 - {\sin}^{2} \alpha = 1 - {\left(\frac{4}{5}\right)}^{2} = \frac{9}{25}$.

Knowing that, $\alpha \in {Q}_{1} , \cos \alpha = + \sqrt{\frac{9}{25}} = + \frac{3}{5}$.

Thus, $\cos \alpha = \frac{3}{5} , w h e r e , \alpha \in {Q}_{1} = \left[0 , \frac{\pi}{2}\right] \subset \left[0 , \pi\right]$.

$\therefore \text{ by definition, } \arccos \left(\frac{3}{5}\right) = \alpha = \arcsin \left(\frac{4}{5}\right)$.

$\Rightarrow \left(\arccos \left(\frac{3}{5}\right) - \arcsin \left(\frac{4}{5}\right)\right) = 0$.

Consequently, $\cos \left(\arccos \left(\frac{3}{5}\right) - \arcsin \left(\frac{4}{5}\right)\right) = \cos 0 = 1$.

$\textcolor{p u r p \le}{\text{Enjoy Maths.!}}$

Jul 14, 2018

$1$

#### Explanation:

$\cos \left(\left(\arccos\right) \left(\frac{3}{5}\right) - a r c \sin e \left(\frac{4}{5}\right)\right)$

:.=cos((arccos(0.6)-arc sine (0.8))#

$\therefore = \cos \left({53}^{\circ} 7 ' 48 ' ' - {53}^{\circ} 7 ' 48 ' '\right)$

$\therefore \cos \left(0\right) = 1$